连接两个范围函数的结果

You can use itertools.chain for this:

from itertools import chain
concatenated = chain(range(30), range(2000, 5002))
for i in concatenated:
...

It works for arbitrary iterables. Note that there's a difference in behavior of range() between Python 2 and 3 that you should know about: in Python 2 range returns a list, and in Python3 an iterator, which is memory-efficient, but not always desirable.

Lists can be concatenated with +, iterators cannot.

range() in Python 2.x returns a list:

>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

xrange() in Python 2.x returns an iterator:

>>> xrange(10)
xrange(10)

And in Python 3 range() also returns an iterator:

>>> r = range(10)
>>> iterator = r.__iter__()
>>> iterator.__next__()
0
>>> iterator.__next__()
1
>>> iterator.__next__()
2

So it is clear that you can not concatenate iterators other by using chain() as the other guy pointed out.

Can be done using list-comprehension.

>>> [i for j in (range(10), range(15, 20)) for i in j]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 16, 17, 18, 19]

Works for your request, but it is a long answer so I will not post it here.

note: can be made into a generator for increased performance:

for x in (i for j in (range(30), range(2000, 5002)) for i in j):
# code

or even into a generator variable.

gen = (i for j in (range(30), range(2000, 5002)) for i in j)
for x in gen:
# code

I like the most simple solutions that are possible (including efficiency). It is not always clear whether the solution is such. Anyway, the range() in Python 3 is a generator. You can wrap it to any construct that does iteration. The list() is capable of construction of a list value from any iterable. The + operator for lists does concatenation. I am using smaller values in the example:

>>> list(range(5))
[0, 1, 2, 3, 4]
>>> list(range(10, 20))
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> list(range(5)) + list(range(10,20))
[0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

This is what range(5) + range(10, 20) exactly did in Python 2.5 -- because range() returned a list.

In Python 3, it is only useful if you really want to construct the list. Otherwise, I recommend the Lev Levitsky's solution with itertools.chain. The documentation also shows the very straightforward implementation:

def chain(*iterables):
# chain('ABC', 'DEF') --> A B C D E F
for it in iterables:
for element in it:
yield element

The solution by Inbar Rose is fine and functionally equivalent. Anyway, my +1 goes to Lev Levitsky and to his argument about using the standard libraries. From The Zen of Python...

In the face of ambiguity, refuse the temptation to guess.

#!python3
import timeit
number = 10000


t = timeit.timeit('''\
for i in itertools.chain(range(30), range(2000, 5002)):
pass
''',
'import itertools', number=number)
print('itertools:', t/number * 1000000, 'microsec/one execution')


t = timeit.timeit('''\
for x in (i for j in (range(30), range(2000, 5002)) for i in j):
pass
''', number=number)
print('generator expression:', t/number * 1000000, 'microsec/one execution')

In my opinion, the itertools.chain is more readable. But what really is important...

itertools: 264.4522138986938 microsec/one execution
generator expression: 785.3081048010291 microsec/one execution

... it is about 3 times faster.

With the help of the extend method, we can concatenate two lists.

>>> a = list(range(1,10))
>>> a.extend(range(100,105))
>>> a
[1, 2, 3, 4, 5, 6, 7, 8, 9, 100, 101, 102, 103, 104]

I came to this question because I was trying to concatenate an unknown number of ranges, that might overlap, and didn't want repeated values in the final iterator. My solution was to use set and the union operator like so:

range1 = range(1,4)
range2 = range(2,6)
concatenated = set.union(set(range1), set(range2)
for i in concatenated:
print(i)

python >= 3.5

You can use iterable unpacking in lists (see PEP 448: Additional Unpacking Generalizations).

If you need a list,

[*range(2, 5), *range(3, 7)]
# [2, 3, 4, 3, 4, 5, 6]

This preserves order and does not remove duplicates. Or, you might want a tuple,

(*range(2, 5), *range(3, 7))
# (2, 3, 4, 3, 4, 5, 6)

... or a set,

# note that this drops duplicates
{*range(2, 5), *range(3, 7)}
# {2, 3, 4, 5, 6}

It also happens to be faster than calling itertools.chain.

from itertools import chain


%timeit list(chain(range(10000), range(5000, 20000)))
%timeit [*range(10000), *range(5000, 20000)]


738 µs ± 10.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
665 µs ± 13.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

The benefit of chain, however, is that you can pass an arbitrary list of ranges.

ranges = [range(2, 5), range(3, 7), ...]
flat = list(chain.from_iterable(ranges))

OTOH, unpacking generalisations haven't been "generalised" to arbitrary sequences, so you will still need to unpack the individual ranges yourself.

You can use list function around range function to make a list LIKE THIS

list(range(3,7))+list(range(2,9))